Let \(V=M_3(\mathbb R)\) and let \[ A=\begin{pmatrix} 1&0&0\\ 0&2&0\\ 0&0&1 \end{pmatrix}. \] Define the linear transformation \(T:V\to V\) by \[ T(X)=\frac12(AX+XA). \] Compute \(\det T\).
Solution
Let \(E_{ij}\) denote the standard matrix unit with a \(1\) in position \((i,j)\) and zeros elsewhere, and write \[ A=\operatorname{diag}(\lambda_1,\lambda_2,\lambda_3) \] with \[ (\lambda_1,\lambda_2,\lambda_3)=(1,2,1). \]
1. For every \(1\leq i,j\leq3\),\[ T(E_{ij}) = \frac{\lambda_i+\lambda_j}{2}E_{ij}. \]
Left multiplication by the diagonal matrix \(A\) scales the \(i\)th row of \(E_{ij}\) by \(\lambda_i\), so \[ AE_{ij}=\lambda_iE_{ij}. \] Right multiplication by \(A\) scales the \(j\)th column by \(\lambda_j\), so \[ E_{ij}A=\lambda_jE_{ij}. \] Therefore \[ T(E_{ij}) = \frac12(AE_{ij}+E_{ij}A) = \frac{\lambda_i+\lambda_j}{2}E_{ij}. \]
2. Relative to the basis\[ \{E_{ij}:1\leq i,j\leq3\}, \]the eigenvalues of \(T\) are\[ 1,\frac32,1, \frac32,2,\frac32, 1,\frac32,1. \]
Step 1 shows that every basis vector \(E_{ij}\) is an eigenvector with eigenvalue \[ \frac{\lambda_i+\lambda_j}{2}. \] Substituting \[ (\lambda_1,\lambda_2,\lambda_3)=(1,2,1) \] for the nine ordered pairs \((i,j)\) gives the displayed list.
3. The determinant is\[ \boxed{\det T=\frac{81}{8}}. \]
Since the matrix of \(T\) in the basis of step 2 is diagonal, its determinant is the product of its nine diagonal entries. Hence \[ \begin{aligned} \det T &= 1^4 \left(\frac32\right)^4 2\\ &= \frac{81}{8}. \end{aligned} \]
4. Q.E.D.
Step 3 gives the requested determinant.